What would the average 5stud Lowball hand be using the Bicycle (A2345) for low and it being seven handed.
If you list out all the hands, including different suits, then the one in the middle is KQJ87. I assume this is what you want. It's actually the median hand.
Here's the spreadsheet that solves this:
<font class="small">Code:</font><hr /><pre>
1317888 np 1 1294336
1098240 1p 2 1272832 1295360
123552 2p 3 1149952 1274880 1296384
54912 set 4 812032 1153024 1277952 1297408
3744 fh 5 1024 816128 1159168 1282048 1298432
624 quads 6 6144 826368 1169408 1287168 1299456
2598960 total 7 21504 846848 1184768 1293312 <font color="red">1300480</font>
2598960 C(52,5) 8 57344 882688 1206272 <font color="red">1300480</font>
<font color="red">1299480</font> C(52,5)/2 9 129024 940032 1234944 1308672
10 258048 1026048 1271808 1317888
11 473088 1148928 <font color="red">1317888</font>
12 811008 <font color="red">1317888 </font>
13 <font color="red">1317888 </font>
</pre><hr />
The first column counts each type of hand.
np: 52*48*44*40*36/5!
1p: 13*6*48*44*40/3!
2p: 13*6*12*6/2*44
set: 13*4*48*44/2
fh: 13*4*12*6
quads: 13*48
The median hand is hand number 1299480 or 1299481. These are both the same hand with different suits, and we see that the median hand is a no-pair. Then we count each no-pair hand from best to worst, and we find that hand 1299480 is a K-high. Then in the next column we count each K-high hand, and we find that hand 1299480 has the second card a Q. We continue like this for each card. The last column starts counting from 1293312, the first KQJ8x. Each column accumulates 4^5*C(n,k) where n are the number of combinations for the remaining cards (number in 3rd column minus 1 = 2 through 12) and k is the number of remaining cards (4,3,2,1,0). 4^5 counts all combinations of suits.
I haven't done the joker case yet. I probably won't be able to do that one without sending you a bill. [img]/images/graemlins/laugh.gif[/img]